Question

We have an expression as $\dfrac{dy}{dx}=\dfrac{x+y+1}{2x+2y+3}$ , its solution is $x+y+\dfrac{k}{3}=c{{e}^{3(x-2y)}}$ what is $k$
1). $1$
2). $2$
3). $5$
4). $4$

Answer 1

First of all assume the term $x+y=t$ then differentiate it with respect to $x$ on both sides and find out the value of $\dfrac{dy}{dx}$ then put the value of $\dfrac{dy}{dx}$ and $t$ in the equation $(1)$ then integrate the whole term and then compare the obtained term with the given solution to check which option is correct among them.

Complete step-by-step solution:
The process of finding the differential coefficient of a function is called differentiation or we can say that differentiation is a process where we find the instantaneous rate of change in function based on one of its variables.
Without requiring addition, multiplication, or quotient equations of differentiation, the derivative or differential coefficient of a function can be calculated directly by the definition of differentiation. The coefficient of $y$ with respect to $x$ is represented by $\dfrac{dy}{dx}$ .
We know that the meaning of $dx$ is the increment in $x$ and it may be positive or negative. Similarly $dy$means the increment in the value of $y$
If there is an increment in the value of $x$ and $y$ in the same direction either both positive or both negative then the value of $\dfrac{dy}{dx}$ is always positive. On the other hand if the increments in $x$ and $y$ are of positive direction that is one positive and other negative then the value of $\dfrac{dy}{dx}$ will be negative.
We have given that $\dfrac{dy}{dx}=\dfrac{x+y+1}{2x+2y+3}$ hence:
Take $2$ in common from denominator we get:
$\Rightarrow \dfrac{dy}{dx}=\dfrac{x+y+1}{2\left( x+y \right)+3}$ mark it as equation $(1)$
Let us assume that $x+y=t$
Now differentiate with respect to $x$ on both sides we will get:
$\Rightarrow 1+\dfrac{dy}{dx}=\dfrac{dt}{dx}$
$\Rightarrow \dfrac{dy}{dx}=\dfrac{dt}{dx}-1$
Put the value of $\dfrac{dy}{dx}$ and $t$ in the equation $(1)$ we get:
$\Rightarrow \dfrac{dt}{dx}-1=\dfrac{t+1}{2t+3}$
$\Rightarrow \dfrac{dt}{dx}=\dfrac{t+1}{2t+3}+1$
$\Rightarrow \dfrac{dt}{dx}=\dfrac{t+1+2t+3}{2t+3}$
$\Rightarrow \dfrac{dt}{dx}=\dfrac{3t+4}{2t+3}$
Taking $t$ terms one side and $dx$ on other side we get:
$\Rightarrow \dfrac{2t+3}{3t+4}dt=dx$
Now integrate both sides we get:
$\Rightarrow \int \dfrac{2t+3}{3t+4}dt=\int dx$
$\Rightarrow \int \dfrac{2t}{3t+4}dt+\int \dfrac{3}{3t+4}dt=x+c$ mark it as equation $(2)$
Now let us integrate the terms one by one:
To integrate $\int \dfrac{2t}{3t+4}dt$
Let $3t+4=u$
Hence we get:
$\Rightarrow 3dt=du$
$\Rightarrow dt=\dfrac{du}{3}$
$= \int \dfrac{2(u-4)}{3u}\cdot \dfrac{du}{3}$ as $t=\dfrac{(u-4)}{3}$
$= \dfrac{2}{9}\int \dfrac{(u-4)}{u}\cdot du$
To integrate $\int \dfrac{3}{3t+4}$
Let $3t+4=u$
Hence we get:
$\Rightarrow 3dt=du$
$\Rightarrow dt=\dfrac{du}{3}$
$= ~\int \dfrac{3}{u}\cdot \dfrac{du}{3}$
$= ~\int \dfrac{1}{u}\cdot du$
Now put these values in $(2)$
$= \dfrac{2}{9}\int \dfrac{(u-4)}{u}\cdot du+\int \dfrac{1}{u}\cdot du=x+c$
$= \dfrac{2}{9}\int 1-\dfrac{4}{u}\cdot du+\int \dfrac{1}{u}\cdot du=x+c$
$= \dfrac{2}{9}\left( u-4\log u \right)+\log u=x+c$
Multiplying the terms we get:
$= \dfrac{2u}{9}-\dfrac{8}{9}\log u+\log u=x+c$
$= \dfrac{2u}{9}+\dfrac{1}{9}\log u=x+c$
Putting the value of $u$ that is $u=3t+4$
$= \dfrac{2\left( 3t+4 \right)}{9}+\dfrac{1}{9}\log \left( 3t+4 \right)=x+c$
Where $t=x+y$
$= \dfrac{2\left( 3\left( x+y \right)+4 \right)}{9}+\dfrac{1}{9}\log \left( 3\left( x+y \right)+4 \right)=x+c$
$= \dfrac{2\left( 3x+3y+4 \right)}{9}+\dfrac{1}{9}\log \left( 3x+3y+4 \right)=x+c$
$= \dfrac{1}{9}\log \left( 3x+3y+4 \right)-\log c=\dfrac{1}{3}\left( x-2y \right)$
Taking $3$ in common we get:
$= \dfrac{3}{9}\log \left( x+y+\dfrac{4}{3} \right)-\log c=\dfrac{1}{3}\left( x-2y \right)$
$= \dfrac{1}{3}\log \left( x+y+\dfrac{4}{3} \right)-\log c=\dfrac{1}{3}\left( x-2y \right)$
$= \log \left( x+y+\dfrac{4}{3} \right)-\log c=\left( x-2y \right)$
Compare this obtained term by the given solution that is $x+y+\dfrac{k}{3}=c{{e}^{3(x-2y)}}$
Hence the value of $k=4$ hence option $(4)$ is correct.

Note: Students must understand the difference between $\dfrac{\delta y}{\delta x}$ and $\dfrac{dy}{dx}$ . Here $\dfrac{\delta y}{\delta x}$ is a fraction with $\delta y$ as a numerator and $\delta x$ as a denominator while $\dfrac{dy}{dx}$ is not a fraction. It is a limiting value of $\dfrac{\delta y}{\delta x}$ . The differential coefficients of those functions which start with $'co'$ like $\text{cosx,cotx,cosecx}$ are always negative.